Solve for x cube root of x+10=8

$\sqrt[3]{x} + 10 = 8$

Move all terms not containing $\sqrt[3]{x}$ to the right side of the equation.

Subtract $10$ from both sides of the equation.

$\sqrt[3]{x} = 8 - 10$

Subtract $10$ from $8$ .

$\sqrt[3]{x} = - 2$

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{x}}^{3} = {(- 2)}^{3}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ .

${(x^{\frac{1}{3}})}^{3} = {(- 2)}^{3}$

Simplify the left side.

Simplify ${(x^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${(x^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{3} \cdot 3} = {(- 2)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$x^{\frac{1}{3} \cdot 3} = {(- 2)}^{3}$

Rewrite the expression.

$x^{1} = {(- 2)}^{3}$

Simplify.

$x = {(- 2)}^{3}$

Simplify the right side.

Raise $- 2$ to the power of $3$ .

$x = - 8$